Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $y = \dfrac{t^2 - 5t + 6}{t + 3} \div \dfrac{7t^2 - 14t}{-2t - 6} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{t^2 - 5t + 6}{t + 3} \times \dfrac{-2t - 6}{7t^2 - 14t} $ First factor the quadratic. $y = \dfrac{(t - 2)(t - 3)}{t + 3} \times \dfrac{-2t - 6}{7t^2 - 14t} $ Then factor out any other terms. $y = \dfrac{(t - 2)(t - 3)}{t + 3} \times \dfrac{-2(t + 3)}{7t(t - 2)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ (t - 2)(t - 3) \times -2(t + 3) } { (t + 3) \times 7t(t - 2) } $ $y = \dfrac{ -2(t - 2)(t - 3)(t + 3)}{ 7t(t + 3)(t - 2)} $ Notice that $(t + 3)$ and $(t - 2)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ -2\cancel{(t - 2)}(t - 3)(t + 3)}{ 7t(t + 3)\cancel{(t - 2)}} $ We are dividing by $t - 2$ , so $t - 2 \neq 0$ Therefore, $t \neq 2$ $y = \dfrac{ -2\cancel{(t - 2)}(t - 3)\cancel{(t + 3)}}{ 7t\cancel{(t + 3)}\cancel{(t - 2)}} $ We are dividing by $t + 3$ , so $t + 3 \neq 0$ Therefore, $t \neq -3$ $y = \dfrac{-2(t - 3)}{7t} ; \space t \neq 2 ; \space t \neq -3 $